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An electron in the hydrogen atom makes a transition from an energy state of principal quantum numbers$n_{i}$ to the $n=2$ state. If the photon emitted has a wavelength of $434 \mathrm{nm},$ what is the value of $n_{\mathrm{i}} ?$

$n_{i}=5$

Chemistry 101

Chapter 7

Quantum Theory and the Electronic Structure of Atoms

Electronic Structure

Carleton College

University of Central Florida

University of Maryland - University College

Brown University

Lectures

04:49

In chemistry and physics, electronic structure is the way the electrons of an atom are arranged in relationship to the nucleus. It is determined by the subshells the electrons are bound to, which are in turn determined by the principal quantum number ("n") and azimuthal quantum number ("l"). The electrons within an atom are attracted to the protons in the nucleus of that atom. The number of electrons bound to the nucleus is equal to the number of protons in the nucleus, which is called the atomic number ("Z"). The electrons are attracted to the nucleus by this mutual attraction and are bound to the nucleus. The electrons within an atom are attracted to each other and this attraction determines the electron configuration. The electron configuration is described by the term symbol, which is the letter used to identify each subshell.

16:45

In physics, the wave–particle duality is the concept that every object or process, no matter how large or how small, behaves as both a wave and a particle. The wave–particle duality is one of the central concepts in quantum mechanics.

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Calculate the wavelength (…

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Okay, So this question is asking that if the collective entity creation emitted as an electron falls from a certain unknown initial quanta to the final in Quanta of n equals two, if that wavelength is 434 centimeters, what is the initial quanta? Okay, so we can't. There's an equation. Client. Sure, energy is equal to plank's constant times frequency. You can expand this equation so that frequency is equal to 2.18 times 10 to the negative 18th jewels divided by Plank's constant, which is 6.63 times 10 to the native, 34th joules per second multiplied by one over the final Quanta. So NF squared, which in this case, we know the final corn is too That's too square, minus the initial Qantas squared. And since this is what we're looking for, this is gonna be our variable. Concede that n I square. Okay, so this is all evil to the frequency so we can find the frequency from the wave like that. They've given us by the relationship between the two. Where frequency? I need to be consistent with my variables. Here. We're frequency. We're just gonna fix that where frequency is equal to speed of light, divided by a wavelength. We know the speed of light is equal to three times 10 to the eighth meters per second. And we know that the wavelength is 434 times 10 to the negative ninth meters. We know that because anything nano meters is always going to be times anything. Man. Oh is gonna be times 10 to the negative night. Okay? And we need this to be in years because the speed of light is in meters per second. So the frequency comes out to be 6.91 times 10 toothy 14th hurts. That makes sense because we cancelled out in the meters here in the meters here when we're left with inverse seconds, which is also able to hurt. Okay, so we can plug this frequency and box this in two. This equation as being the frequency here, and then we can solve for the initial quanta. Now, this just doing algebra. You're going to divide this by this. Okay, then you're gonna divide this 6.91 time. Send the 14th hurts by whatever this answer is this thing divided by this thing. Okay, then you so skipping that step, we would get 0.21 equals. We're gonna go ahead and say instead of 14 I'm gonna say 0.25 because we have decimals and I don't know what 0.21 is infraction minus one over the initial quantum, which we know as an I square. So we want the variable by itself so we can go ahead and say one over the initial Qantas Square is going to be equal to 0.4 We can go ahead and take the square root of both sides. Do you get one over the initial quanta two equals 0.2. And when we find a inverse so basically real multiply initial quanta this variable by both sides that this cancels in this cancels we will get one is equal to zero point to time is variable. And when we divide, but is there a point to on both sides? One divided by 0.2 is five. So this means that for the just based off of the fact that we know which energy level which quanta it ended at as because it Final Quanta being too. And we know that the wavelength of the electromagnetic radiation emitted appear plugging all of this into this equation and solving for the frequency or some for the first into playing it all into this equation, we can find that the Quanta It's the energized state for quanta that the electron fell from on its way to to is Quantas Energy level number five. Thank you very much.

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